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Homework Help: Coefficient of Friction Question.

  1. May 18, 2008 #1
    Hey, I would like to check that my working for the following question is along the right lines. I have an exam this coming Wednesday and there are a few really common questions that I think I should do some work on. As always any help is much appreciated! :smile:

    The Question

    A parcel of weight 10N lies on a rough plane inclined at an angle of 30 degrees to the horizontal. A horizontal force of magnitude P newtons acts on the parcel as shown in the attachment (Bottom of post). The parcel is in equilibrium and on the point of slipping up the plane. The normal reaction of the plane on the parcel is 18N. The coefficient of friction between the parcel and the plane is [itex]\mu[/itex]. Find:

    (a) The value of P

    (b) The value of [itex]\mu[/itex]

    The horizontal force is removed.

    (c) Determine whether or not the parcel moves.

    My Attempt

    (a) Resolve perpendicular to the plane:

    [itex]18 - 10\cos30 - P\sin 30[/itex]

    [itex]18 = 10\cos30 + P\sin 30[/itex]

    [itex]18 - 5\sqrt3 = \frac{P}{2}[/itex]

    [itex]2(18 - 5\sqrt3) = P = 18.7N[/itex]

    (b) Resolve along the plane.

    [itex]F + 10\sin30 = 18.7\cos30[/itex]

    [itex]5F = 16.20[/itex]

    [itex]F = 3.24[/itex]

    Now I just plug in the answer from (a) with this one from (b)

    [itex]F=\mu R[/itex]

    [itex]3.24 = 18\mu[/itex]

    [itex]\mu = 0.18[/itex]

    I am still working on (c) but could someone please have a look at how I am doing so far. I may have skipped a few steps, but I am just using the fact that when resolving perpendicular you can always use cos, and whenever you resolve parallel you can use sin, when dealing with vectors int his course on an inclined plane.

    Thank you.


    Attached Files:

  2. jcsd
  3. May 18, 2008 #2

    Doc Al

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    Staff: Mentor

    I didn't check your arithmetic, but your method is correct.
  4. May 18, 2008 #3
    I think you have made an arithmetic mistake in part (b) - the second line of working for (b) is inconsistent with the first line.
  5. May 18, 2008 #4
    For (c), consider the maximum value which friction can take, remembering that now that P has been removed, the normal reaction will change. Compare this to the component of weight down the slope.
  6. May 18, 2008 #5

    Doc Al

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    Staff: Mentor

    nokia8650 is correct--you made an arithmetic mistake in part (b).
  7. May 18, 2008 #6
    [itex]F + 10\sin30 = 18.7\cos30 [/itex]
    [itex]5F = 16.20[/itex]

    Ah right!

    [itex]F + 10\sin30 = 18.7\cos30[/itex]
    [itex]F + 5 = 16.20[/itex]
    [itex]F = 11.20[/itex]

    [itex]\mu = 0.62[/itex]

    Thanks for that, silly mistake. I'll have a look at your advice in a minute or two nokia. Thank you both Nokia and Doc.

    _Mayday_ :approve:
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